![]() ![]() An extra phone running Apps I use watching TV or working at my laptop. Great idea for a spare phone at home that does not need mobility as a feature - that is exactly why I would want to do it. If I connect the red and black wires of the Nook to this Almac device, will it power on the Nook, without a battery and without connecting it to its own wall charger? :) I've found a small device named "almac combo of 5v 1A Micro Usb input for 3.7v 18650 Lithium Battery Charger With Protection Module, best for robotics DIY kit" which seems to be doing the same thing. The method you've outlined above gives the current to the device without using a battery. The latest battery I tried is an Ipad one, and with the Ipad battery alone, the Nook doesn't power on, and when I connect the wall charger and turn it on, the Nook display turns on for half a second and then turns off and then nothing happens. What I did is I cut the wires coming from the dead battery and attached these two (red and black) wires to a different battery with the same specs (3.7 volts) so that on one side the wires are connected to the new battery and on the other side, they are connected to the tablet like they were with the original battery. I've tried attaching different batteries but none seem to work. I have an old Nook HD+ tablet the battery of which died. As for the procedure itself and instalation of it all, that is for everyone to decide as it fits. This will totally suffice to get what you intended here. And the least complicated way to achieve the goal. The procedure in this post by itself is technically correct. As to the link between the other two pins of the battery connections and the values and dependancies. I Included them in to consideration, but did not mention them. In fact there are many facts surrounding this area and to consider. DISCLAIMER: This is a direct instruction for thisws case, this post. More safety issues and not intended for such low power loads. Ussualy a very powerful power supplies eve would not work in this way. ![]() WARNING! Do not use a power supply with too high values or to much power. In any case of source it should match the specified voltage of the device as much as possible, if not rather exactly in any way. THAN RECALCULATE THE VALUE OF A RESISTOR NEEDED. USE a charger or DC power supply with 2A or higher. HOWEVER! This current is not sufficient for most of the smart phones nowadays. Which is a drop we need to get 3,8V of 5V source. Use a resistor lin linear positive wire and with the value calculated via Ohms law R=U/I. But lower than nominal is! As in this case voltage needs to be lowered from 5V to 3.8V. At the same time the source must also be capable of at least the current needed by the device plus 5% higher of that value. Use resistor with value calculated to achieve desired delivery voltage. ![]()
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